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Aditya Nijampurkar Grade: 12
        A steel ball of radius R=20cm and maass =2kg is rotating about a horizontal diameter with angular velocity w naught=50 rad/sec. this rotating ball is dropped onto a rough horizontal floor and falls freely through a height of h=1.25 m. the coefficient of restitution is e=1.0 and coefficient of friction between ball and the floor is mu=0.3. calculate 

(1)distance between points of first and second impact of the ball with the floor.
(2)loss of energy due to friction
(g=10m/s^2)
7 years ago

Answers : (1)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

Dear Aditya Nijampurkar,


Ans:- See the force of friction will give it a horizontal velocity but unless the time of contact is not given we just can' t find the horizontal velocity of the ball. Hence we can't get the correct answer. Now if we consider that the time of contact is so high that the ball has started pure rolling in the forward direction, then the required eq are,


W=Wo - βt.................(1)


V=at..........................(2)


 and for pure rolling Wr=V.................(3)


and a=F/m,.................(4)


β=Fr/I where I = moment of inertia=2/5 M r²................(5)


solving we get


V=2/7 W0 r = 20/7 m/s


Again the loss of KE = Initial KE - Final total energy


                            =1/5 M (Wo r)²-2/35M (Wo r)²


                             =1/7M(Wo r)²


                               =200/7  J(ans)


 


 


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All the best Aditya Nijampurkar !!!


 



Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

7 years ago
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